Open ended pipe Harmonics Mastering Physics Question

AI Thread Summary
The discussion centers on determining the highest harmonic frequency for a closed-end pipe measuring 45.0 cm, using the speed of sound at 344 m/s. The correct formula for harmonics in a closed pipe is f_n = (2n-1)(v/4L), leading to the conclusion that for n = 52, the harmonic number is actually 103. The user initially calculated a frequency of 52.5 but struggled with the Mastering Physics system, which rejected their answers. Ultimately, the correct highest harmonic below 20,000 Hz is confirmed to be 103, clarifying the relationship between n and the harmonic number. The discussion concludes with an acknowledgment of the correct approach to the problem.
TFM
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[SOLVED] Open ended pipe Harmonics Mastering Physics Question

Homework Statement



Consider a pipe 45.0cm long if the pipe is open at both ends. Use v = 344m/s.
Now pipe is closed at one end.

What is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20 Hz to 20000 Hz?

Homework Equations



f_n = (2n-1)\frac{v}{4L}

The Attempt at a Solution



I have an answer that works, but masteringphysics doesn't accept. I first rearranged the equation to give me:

(2n-1) = \frac{f_n * 4L}{v}

then:

2n = (\frac{f_n * 4L}{v})+1

and finally:

n = ((\frac{f_n * 4L}{v})+1)/2

inserting the values gives 52.5 so I inserted 52 as the answer. wrong, I have tried 51-54, all wrong. so I thought tpo go backwards, using:

(2n-1) = \frac{f_n * 4L}{v}

and inserting values, to find the value which is the closest to 20000, buit under it - guess what, the value that came out:

52!

Any ideas

TFM
 
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The harmonics of a pipe closed at one end are all odd. For n = 2 you have the 3rd harmonic. For n=52 what harmonic do you have?
 
It will be the 53rd Harmonic. The trouble is, I have put 53 in, and it says its the wrong answer!
 
TFM said:
It will be the 53rd Harmonic. The trouble is, I have put 53 in, and it says its the wrong answer!

Sorry that third harmonic was a bad example. The harmonics are given by 2n-1. So if n is 52 what is the harmonic. An easier way to have thought about it would to have solved for:

f_n = \frac{nv}{4L}

for n = 1, 3, 5,...
 
Using:

f_n = \frac{nv}{4L}

and using n = 103,

I get a frequency of 19684, which is the first odd number below 20000. would this be the harmonic number?

TFM
 
TFM said:
I get a frequency of 19684, which is the first odd number below 20000. would this be the harmonic number?

TFM

Yes n is the harmonic number.
 
Success! n = 103.

IOne thing does bother me slightly - where does my orginal answer of 52 fit in?

TFM
 
2n - 1 is just another way of saying n = 1, 3, 5, ... . So if you stick n = 52 into 2n - 1 you get 103.
 
Last edited:
That makes sense.

Thanks,

TFM
 
  • #10
TFM said:
That makes sense.

Thanks,

TFM

What I was originally aiming at was for you to put the n = 52 into that equation and get 103 but I used a stupid example which probably mislead you slightly. :smile:
 
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